Integrand size = 24, antiderivative size = 118 \[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=-\frac {3 \sqrt {\arctan (a x)}}{32 a^4 c^3}+\frac {x^4 \sqrt {\arctan (a x)}}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{64 a^4 c^3}+\frac {\sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{16 a^4 c^3} \]
-1/128*FresnelC(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^4 /c^3+1/16*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^4/c^3-3/32*arc tan(a*x)^(1/2)/a^4/c^3+1/4*x^4*arctan(a*x)^(1/2)/c^3/(a^2*x^2+1)^2
Result contains complex when optimal does not.
Time = 0.56 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.95 \[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {-10 \sqrt {2 \pi } \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+80 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\frac {\frac {64 \left (-3-6 a^2 x^2+5 a^4 x^4\right ) \arctan (a x)}{\left (1+a^2 x^2\right )^2}-12 i \sqrt {2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-2 i \arctan (a x)\right )+12 i \sqrt {2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},2 i \arctan (a x)\right )+3 i \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-4 i \arctan (a x)\right )-3 i \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},4 i \arctan (a x)\right )}{\sqrt {\arctan (a x)}}}{2048 a^4 c^3} \]
(-10*Sqrt[2*Pi]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]] + 80*Sqrt[Pi]*Fre snelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + ((64*(-3 - 6*a^2*x^2 + 5*a^4*x^4)* ArcTan[a*x])/(1 + a^2*x^2)^2 - (12*I)*Sqrt[2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma [1/2, (-2*I)*ArcTan[a*x]] + (12*I)*Sqrt[2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (2*I)*ArcTan[a*x]] + (3*I)*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-4*I)*ArcTan [a*x]] - (3*I)*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (4*I)*ArcTan[a*x]])/Sqrt[Arc Tan[a*x]])/(2048*a^4*c^3)
Time = 0.47 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5479, 27, 5505, 3042, 3793, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (a^2 c x^2+c\right )^3} \, dx\) |
\(\Big \downarrow \) 5479 |
\(\displaystyle \frac {x^4 \sqrt {\arctan (a x)}}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {1}{8} a \int \frac {x^4}{c^3 \left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^4 \sqrt {\arctan (a x)}}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {a \int \frac {x^4}{\left (a^2 x^2+1\right )^3 \sqrt {\arctan (a x)}}dx}{8 c^3}\) |
\(\Big \downarrow \) 5505 |
\(\displaystyle \frac {x^4 \sqrt {\arctan (a x)}}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {\int \frac {a^4 x^4}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}d\arctan (a x)}{8 a^4 c^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {x^4 \sqrt {\arctan (a x)}}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {\int \frac {\sin (\arctan (a x))^4}{\sqrt {\arctan (a x)}}d\arctan (a x)}{8 a^4 c^3}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \frac {x^4 \sqrt {\arctan (a x)}}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {\int \left (-\frac {\cos (2 \arctan (a x))}{2 \sqrt {\arctan (a x)}}+\frac {\cos (4 \arctan (a x))}{8 \sqrt {\arctan (a x)}}+\frac {3}{8 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{8 a^4 c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^4 \sqrt {\arctan (a x)}}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {\frac {1}{8} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+\frac {3}{4} \sqrt {\arctan (a x)}}{8 a^4 c^3}\) |
(x^4*Sqrt[ArcTan[a*x]])/(4*c^3*(1 + a^2*x^2)^2) - ((3*Sqrt[ArcTan[a*x]])/4 + (Sqrt[Pi/2]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/8 - (Sqrt[Pi]*Fre snelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/2)/(8*a^4*c^3)
3.8.16.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(d*f*(m + 1))), x] - Simp[b*c*(p/(f*(m + 1))) Int[(f*x) ^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(q_), x_Symbol] :> Simp[d^q/c^(m + 1) Subst[Int[(a + b*x)^p*(Sin[x]^m/ Cos[x]^(m + 2*(q + 1))), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p }, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q ] || GtQ[d, 0])
Time = 6.40 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79
method | result | size |
default | \(-\frac {\operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \sqrt {2}\, \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }+16 \cos \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-8 \,\operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }-4 \cos \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )}{128 c^{3} a^{4} \sqrt {\arctan \left (a x \right )}}\) | \(93\) |
-1/128/c^3/a^4/arctan(a*x)^(1/2)*(FresnelC(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^ (1/2))*2^(1/2)*arctan(a*x)^(1/2)*Pi^(1/2)+16*cos(2*arctan(a*x))*arctan(a*x )-8*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*arctan(a*x)^(1/2)*Pi^(1/2)-4*co s(4*arctan(a*x))*arctan(a*x))
Exception generated. \[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\frac {\int \frac {x^{3} \sqrt {\operatorname {atan}{\left (a x \right )}}}{a^{6} x^{6} + 3 a^{4} x^{4} + 3 a^{2} x^{2} + 1}\, dx}{c^{3}} \]
Exception generated. \[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\int { \frac {x^{3} \sqrt {\arctan \left (a x\right )}}{{\left (a^{2} c x^{2} + c\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^3} \, dx=\int \frac {x^3\,\sqrt {\mathrm {atan}\left (a\,x\right )}}{{\left (c\,a^2\,x^2+c\right )}^3} \,d x \]